What is the integral of (x^2)(4-x^3)^4 ?

The integral of (x^2)(4-x^3)^4 is 1/15 (-4+x^3)^5+C

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integral of (x^2)(4-x^3)^4

\int (x^{2}) (4 - x^{3})^{4} d x

\frac{1}{15} (- 4 + x^{3})^{5} + C

Solution steps

\int x^{2} (4 - x^{3})^{4} d x

Factor

4 - x^{3} : - (- 4 + x^{3})

= \int x^{2} (- (- 4 + x^{3}))^{4} d x

(- (- 4 + x^{3}))^{4} = (- 4 + x^{3})^{4}

= \int x^{2} (- 4 + x^{3})^{4} d x

Apply u-substitution

: \int \frac{u ^{4}}{3} d u

= \int \frac{u ^{4}}{3} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{3} \cdot \int u^{4} d u

Apply the Power Rule

: \frac{u ^{5}}{5}

= \frac{1}{3} \cdot \frac{u ^{5}}{5}

Substitute back

u = - 4 + x^{3}

= \frac{1}{3} \cdot \frac{( - 4 + x ^{3} ) ^{5}}{5}

Simplify

\frac{1}{3} \cdot \frac{( - 4 + x ^{3} ) ^{5}}{5} : \frac{1}{15} (- 4 + x^{3})^{5}

= \frac{1}{15} (- 4 + x^{3})^{5}

Add a constant to the solution

= \frac{1}{15} (- 4 + x^{3})^{5} + C

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What is the integral of (x^2)(4-x^3)^4 ?
The integral of (x^2)(4-x^3)^4 is 1/15 (-4+x^3)^5+C